$\underset{n\to \infty }{\lim }\frac{(3n+2{)}^2-(1-2n{)}^2}{(2n-1{)}^2}=\underset{n\to \infty }{\lim }\frac{9n^2+12n+4-(1-4n+4n^2)}{4n^2-4n+1}=$

$=\underset{n\to \infty }{\lim }\frac{5n^2+16n+3}{4n^2-4n+1}=\underset{n\to \infty }{\lim }\frac{\frac{5n^2}{n^2}+\frac{16n}{n^2}+\frac{3}{n^2}}{\frac{4n^2}{n^2}-\frac{4n}{n^2}+\frac{1}{n^2}}=\underset{n\to \infty }{\lim }\frac{5+\frac{16}{n}+\frac{3}{n^2}}{4-\frac{4}{n}+\frac{1}{n^2}}=\frac{5}{4}$

Odp.: Granica jest równa $\frac{5}{4}$.